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This question already has an answer here:

The following formula is given in our lab manual: $$ T = 2 \pi \sqrt{\frac{L}{g}} \left( 1 + \frac{1}{4}\sin^2 \frac{\theta}{2} + \frac{9}{64}\sin^4 \frac{\theta}{2}+\cdots \right) $$ for the period of a simple pendulum with length $L$ under the influence of gravitational acceleration with magnitude $g$. We ignore friction. I understand the solution of Newton's Equation via the small angle approximation $\sin \theta \approxeq \theta$ and why that gives the usual elementary formula $T = 2 \pi \sqrt{\frac{L}{g}}$. What I don't understand is how one derives the above formula for higher-order corrections to the period. I think the $\theta$ in the formula is the maximum angle to which the pendulum in question is swinging.

Any insight into the origin or derivation of this formula is appreciated.

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marked as duplicate by Bill N, John Rennie newtonian-mechanics Jan 23 at 6:16

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ See the section “Legendre polynomial solution for the elliptic integral” at en.wikipedia.org/wiki/… $\endgroup$ – G. Smith Jan 22 at 17:24
  • $\begingroup$ BTW, there's a much nicer formula if you're doing this on a computer, using the AGM, the arithmetic-geometric mean, which converges very quickly. $\endgroup$ – PM 2Ring Jan 22 at 17:31
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    $\begingroup$ Possible duplicate of Period of a pendulum See the answer here. The solution is found in different textbooks, too. e.g., Engineering Dynamics by Beer & Johnston, , Classical Dynamics by Marion, This derivation is also a problem in Classical Mchanics by Taylor, so OP might be looking for that .... $\endgroup$ – Bill N Jan 22 at 18:30
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    $\begingroup$ That answer doesn’t explain how to get the series. $\endgroup$ – G. Smith Jan 22 at 18:43
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    $\begingroup$ @BillN I don't spend much time over here in Physics, but that seems pretty far from a direct duplicate. If I was searching for the series formula for the period I don't think that Q and A would be directly helpful. $\endgroup$ – James S. Cook Jan 23 at 1:14
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If we take the zero of gravitational potential energy to be at the pivot, then by conservation of energy,

$$\frac{1}{2}m(L\dot{\theta})^2-m g L\cos{\theta}=-m g L\cos{\theta_0}$$

where $\theta_0$ is the maximum angle (where the kinetic energy is zero). This gives the differential equation

$$\frac{d\theta}{dt}=\sqrt{\frac{2g}{L}(\cos{\theta}-\cos{\theta_0}})$$

which gives the period as four times the time to go from 0 to $\theta_0$:

$$T=4\sqrt\frac{L}{2g}\int_0^{\theta_0}\frac{d\theta}{\sqrt{\cos{\theta}-\cos{\theta_0}}}.$$

Using the trigonometric identity

$$\cos\theta=1-2\sin^2{\frac{\theta}{2}}$$

this becomes

$$T=2\sqrt\frac{L}{g}\int_0^{\theta_0}\frac{d\theta}{\sqrt{\sin^2{\frac{\theta_0}{2}}-\sin^2{\frac{\theta}{2}}}}.$$

With the change of integration variable to $u$ where

$$\sin{u}=\frac{\sin{\frac{\theta}{2}}}{\sin{\frac{\theta_0}{2}}}$$

this becomes

$$T=4\sqrt\frac{L}{g}\int_0^{\pi/2}\frac{du}{\sqrt{1-k^2\sin^2{u}}}$$

where

$$k=\sin\frac{\theta_0}{2}.$$

Now use a Taylor expansion and integrate term-by-term to get the result you were given:

$$\begin{align} T&=4\sqrt\frac{L}{g}\int_0^{\pi/2}du\;\left(1+\frac{1}{2}k^2\sin^2{u}+\frac{3}{8}k^4\sin^4{u}+...\right)\\ &=4\sqrt\frac{L}{g}\left(\frac{\pi}{2}+\frac{1}{2}k^2\frac{\pi}{4}+\frac{3}{8}k^4\frac{3\pi}{16}+...\right) \\ &=2\pi\sqrt\frac{L}{g}\left(1+\frac{1}{4}k^2+\frac{9}{64}k^4+...\right). \end{align}$$

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  • $\begingroup$ I think you'd enjoy The AGM Simple Pendulum by Mark B. Villarino. $\endgroup$ – PM 2Ring Jan 22 at 18:45
  • $\begingroup$ Thanks for this answer. Exactly the sort of detail I had hoped for. $\endgroup$ – James S. Cook Jan 22 at 21:21

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